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-(16x^2-35x+19)=0
We get rid of parentheses
-16x^2+35x-19=0
a = -16; b = 35; c = -19;
Δ = b2-4ac
Δ = 352-4·(-16)·(-19)
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{9}=3$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(35)-3}{2*-16}=\frac{-38}{-32} =1+3/16 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(35)+3}{2*-16}=\frac{-32}{-32} =1 $
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